Let $g(x)=3x^5-30x^4$. For what values of $x$ does the graph of $g$ have a point of inflection? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=0$ (Choice B) B $x=6$ (Choice C) C $x=10$ (Choice D) D $g$ has no points of inflection.
Explanation: We can find the inflection points of the graph of $g$ by looking for the intervals where its second derivative $g''$ is positive/negative. This analysis is very similar to finding minimum/maximum points, only instead of analyzing $g'$, we are analyzing $g''$. The second derivative of $g$ is $g''(x)=60x^2(x-6)$. $g''(x)=0$ for $x=0,6$. Since $g''$ is a polynomial, it's defined for all real numbers. Therefore, our possible inflection points are $x=0$ and $x=6$. Our possible inflection points divide the number line into three intervals: $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $x< \llap{-}2$ $\llap{-}2<x<1$ $x>1$ Let's evaluate $g''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g''(x)$ Verdict $x<0$ $x=-1$ $g''(-1)=-420<0$ $g$ is concave down $\cap$ $0<x<6$ $x=1$ $g''(1)=-300<0$ $g$ is concave down $\cap$ $x>6$ $x=7$ $g''(7)=2940>0$ $g$ is concave up $\cup$ We can see that the graph of $g$ changes concavity at $x=6$. In conclusion, $g$ has a point of inflection at $x=6$.